Given an array of strings arr[], the task is to find the smallest string which contains all the characters of the given array of strings.
Examples:
Input: arr[] = {“your”, “you”, “or”, “yo”}
Output: ruyo
Explanation: The string “ruyo” is the smallest string which contains all the characters that are used across all the strings of the given array.Input: arr[] = {“abm”, “bmt”, “cd”, “tca”}
Output: abctdm
Approach: This problem can be solved by using the Set Data Structure. Set has the capability to remove duplicates, which is needed in this problem in order to minimize the string size. Add all the characters in the set from all the strings in the array arr[] and form a string containing all the characters remaining in the set, which is the required answer.
Below is the implementation of the above approach.
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;string minSubstr(vector<string> s){ // Stores the concatenated string // of all the given strings string str = ""; // Loop to iterate through all // the given strings for (int i = 0; i < s.size(); i++) { str += s[i]; } // Set to store the characters unordered_set<char> set; // Loop to iterate over all // the characters in str for (int i = 0; i < str.length(); i++) { set.insert(str[i]); } string res = ""; // Loop to iterate over the set for (auto itr = set.begin(); itr != set.end(); itr++) { res = res + (*itr); } // Return Answer return res;}// Driver Codeint main(){ vector<string> arr = {"your", "you", "or", "yo"}; cout << (minSubstr(arr)); return 0;}// This code is contributed by Potta Lokesh |
Java
import java.util.*;public class GfG { public static String minSubstr(String s[]) { // Stores the concatenated string // of all the given strings String str = ""; // Loop to iterate through all // the given strings for (int i = 0; i < s.length; i++) { str += s[i]; } // Set to store the characters Set<Character> set = new HashSet<Character>(); // Loop to iterate over all // the characters in str for (int i = 0; i < str.length(); i++) { set.add(str.charAt(i)); } // Stores the required answer String res = ""; Iterator<Character> itr = set.iterator(); // Loop to iterate over the set while (itr.hasNext()) { res += itr.next(); } // Return Answer return res; } // Driver Code public static void main(String[] args) { String arr[] = new String[] { "your", "you", "or", "yo" }; System.out.println(minSubstr(arr)); }} |
Python3
# Python code for the above approachdef minSubstr(s): # Stores the concatenated string # of all the given strings str = "" # Loop to iterate through all # the given strings for i in range(len(s)): str += s[i] # Set to store the characters _set = set() # Loop to iterate over all # the characters in str for i in range(len(str)): _set.add(str[i]) # Stores the required answer res = "" # Loop to iterate over the set for itr in _set: res += itr # Return Answer return res# Driver Codearr = ["your", "you", "or", "yo"]print(minSubstr(arr))# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ public static string minSubstr(string []s) { // Stores the concatenated string // of all the given strings string str = ""; // Loop to iterate through all // the given strings for (int i = 0; i < s.Length; i++) { str += s[i]; } // Set to store the characters HashSet<char> set = new HashSet<char>(); // Loop to iterate over all // the characters in str for (int i = 0; i < str.Length; i++) { set.Add(str[i]); } // Stores the required answer String res = ""; // Loop to iterate over the set foreach(char i in set) { res += i; } // Return Answer return res; } // Driver Code public static void Main() { string []arr = { "your", "you", "or", "yo" }; Console.WriteLine(minSubstr(arr)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach function minSubstr(s) { // Stores the concatenated string // of all the given strings let str = ""; // Loop to iterate through all // the given strings for (let i = 0; i < s.length; i++) { str += s[i]; } // Set to store the characters let set = new Set(); // Loop to iterate over all // the characters in str for (let i = 0; i < str.length; i++) { set.add(str[i]); } // Stores the required answer let res = ""; // Loop to iterate over the set for (let itr of set) { res += itr; } // Return Answer return res; } // Driver Code let arr = ["your", "you", "or", "yo"]; document.write(minSubstr(arr));// This code is contributed by Potta Lokesh </script> |
Output
ruoy
Time Complexity: O(N*M), where M is the average length of strings in the given array
Auxiliary Space: O(N) because extra space for string str is being used
Approach #2: This problem can also be solved by using the Map Data Structure. Map stores all the characters present in the string with their occurrence. After iterating on the map we will get the all unique characters.
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;string minSubstr(vector<string> s){ // Stores the concatenated string // of all the given strings string str = ""; // Loop to iterate through all // the given strings for (int i = 0; i < s.size(); i++) { str += s[i]; } // map to store the characters with frequency unordered_map<char, int> mp; // Loop to iterate over all // the characters in str for (int i = 0; i < str.length(); i++) { mp[str[i]]++; } string res = ""; // Loop to iterate over the map for (auto it : mp) { res += it.first; } // Return Answer return res;}// Driver Codeint main(){ vector<string> arr = { "your", "you", "or", "yo" }; cout << (minSubstr(arr)); return 0;}// This code is contributed by Prasad Kandekar(prasad264) |
Java
// Java code for the above approachimport java.util.*;class GFG { public static String minSubstr(List<String> s) { // Stores the concatenated string // of all the given strings String str = ""; // Loop to iterate through all // the given strings for (String x : s) { str += x; } // map to store the characters with frequency Map<Character, Integer> mp = new HashMap<>(); // Loop to iterate over all // the characters in str for (int i = 0; i < str.length(); i++) { char c = str.charAt(i); if (mp.containsKey(c)) { mp.put(c, mp.get(c) + 1); } else { mp.put(c, 1); } } StringBuilder res = new StringBuilder(); // Loop to iterate over the map for (Map.Entry<Character, Integer> entry : mp.entrySet()) { res.append(entry.getKey()); } // Return Answer return res.toString(); } // Driver Code public static void main(String[] args) { List<String> arr = Arrays.asList("your", "you", "or", "yo"); System.out.println(minSubstr(arr)); }}// This code is contributed by Prasad Kandekar(prasad264) |
Python3
# Python code for the above approachdef min_substr(s): # Stores the concatenated string # of all the given strings str = "" # Loop to iterate through all # the given strings for i in range(len(s)): str += s[i] # dictionary to store the characters with frequency mp = {} # Loop to iterate over all # the characters in str for i in range(len(str)): if str[i] in mp: mp[str[i]] += 1 else: mp[str[i]] = 1 res = "" # Loop to iterate over the map for key in mp: res += key # Return Answer return res# Driver Codearr = ["your", "you", "or", "yo"]print(min_substr(arr))# This code is contributed by karthik |
C#
// C# code for the above approachusing System;using System.Collections.Generic;public class GFG { static string MinSubstr(List<string> s) { // Stores the concatenated string // of all the given strings string str = ""; // Loop to iterate through all // the given strings for (int i = 0; i < s.Count; i++) { str += s[i]; } // dictionary to store the characters with frequency Dictionary<char, int> mp = new Dictionary<char, int>(); // Loop to iterate over all // the characters in str for (int i = 0; i < str.Length; i++) { if (mp.ContainsKey(str[i])) { mp[str[i]]++; } else { mp[str[i]] = 1; } } string res = ""; // Loop to iterate over the map foreach(var item in mp) { res += item.Key; } // Return Answer return res; } // Driver Code static public void Main(string[] args) { List<string> arr = new List<string>() { "your", "you", "or", "yo" }; Console.WriteLine(MinSubstr(arr)); }}// This code is contributed by Prasad Kandekar(prasad264) |
Javascript
// JavaScript code for the above approachfunction minSubstr(s) { // Stores the concatenated string // of all the given strings var str = ""; // Loop to iterate through // all the given strings for (var i = 0; i < s.length; i++) { str += s[i]; } // map to store the characters with frequency var mp = new Map(); // Loop to iterate over all // the characters in str for (var i = 0; i < str.length; i++) { if (mp.has(str[i])) { mp.set(str[i], mp.get(str[i]) + 1); } else { mp.set(str[i], 1); } } var res = ""; // Loop to iterate over the map for (var [key, value] of mp) { res += key; } // Return Answer return res;}// Driver Codevar arr = ["your", "you", "or", "yo"];console.log(minSubstr(arr));// This code is contributed by Prasad Kandekar(prasad264) |
Output:
ruoy
Complexity analysis:
Time Complexity: O(N*M), where M is the average length of strings in the given array
Auxiliary Space: O(N) because extra space for string str and unordered_map are being used
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
